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\begin{document}

\centerline{\bf Fall 2018 Special and General Relativity (PHZ 4601/560)} 
% \bigskip \vskip -20pt

\centerline{\bf Final Solutions} 
% \bigskip \vskip -20pt

\begin{enumerate}

\item{\bf Special Relativity: Electromagnetic invariants.}
\smallskip

\begin{enumerate}
\item 
$$ \left(E^{\alpha\beta}\right)=\left(\matrix{
    0  & -b^3&  b^2&  e^1\cr
    b^3&  0  & -b^1&  e^2\cr
   -b^2&  b^1&  0  &  e^3\cr
   -e^1& -e^2& -e^3&  0  }\right)~~\Rightarrow~~
   \left(E_{\alpha\beta}\right)=\left(\matrix{
    0  & -b^3&  b^2& -e^1\cr
    b^3&  0  & -b^1& -e^2\cr
   -b^2&  b^1&  0  & -e^3\cr 
   +e^1& +e^2& +e^3&  0  }\right)
\,.$$ 
Therefore,
$$ E_{\alpha\beta}\,E^{\alpha\beta} = 2\,\left(\vec{b}^{\,2}
   - \vec{e}^{\,2} \right) \ . $$  \medskip

\item $E_{12}=\epsilon_{1234}\,E^{34}=-E^{34}=+e^3,\qquad
       E_{13}=\epsilon_{1324}\,E^{24}=+E^{24}=-e^1$,

      $E_{13}=\epsilon_{1324}\,E^{24}=+E^{24}=-e^1$, and so on implies
$$ \left(^*\!E_{\alpha\beta}\right)=\left(\matrix{
    0  & -e^3&  e^2&  b^1\cr
    e^3&  0  & -e^1&  b^2\cr
   -e^2&  e^1&  0  &  b^3\cr
   -b^1& -b^2& -b^3&  0  }\right) \,.$$ 

\item Therefore,
$$ ^*\!E_{\alpha\beta}\,E^{\alpha\beta} = 4\,\vec{b}\cdot\vec{e}\ .$$ 
% \medskip
\end{enumerate}
\bigskip

\item{\bf Schwarzschild effective potential:}

\begin{enumerate}
\item We have $h=r_e\,v_e$ where $r_e$ is the average distance of 
the earth from the sun and $v_e$ the average velocity of the earth 
perpendicular to $r_e$. Approximating the orbit of the earth by a
circle,
$$ v_e = \frac{2\pi\,r_e}{T} = 0.9962\times 10^{-4}\,, $$
where the period $T$ is one earth year and we use gravitational units 
with the mass of the sun as unit mass. Multiplying with $r_e$, we 
obtain 
$$ h = 9962\,[m_s]\,.$$
Second solution for (a): The values of $h$ can also be found from
the minimum of $V_{GR}(r)$, which corresponds to the orbit of a
circle. Then
$$ 0 = V'_{GR}(r) = \frac{d~}{dr}\,V_{GR}(r)= -\frac{2\,h^2}{r^3}
     + \frac{2}{r^2} + \frac{6\,h^2}{r^4}~~\Rightarrow~~
   h^2 = \left(\frac{2}{r^2}\right)/\left(\frac{2}{r^3}-
         \frac{6}{r^4}\right)\,,$$
and with $r=r_e=10^8\,[m_s]$ we get 
$$ h\approx r_e = 10^4\,[m_s]\,.$$
The difference between the two results comes from the approximation
for $T$.

\begin{figure}[th]
\centerline{\hbox{\psfig{figure=figures/VGR100000N.eps,width=10cm}}}
Newton's classical region. % \end{figure} \begin{figure}[tb]
\centerline{\hbox{\psfig{figure=figures/VGR100000S.eps,width=10cm}}}
Schwarzschild singularity region.  \end{figure}

\item The extrema of the Schwarzschild effective potential follow
from 
$$ 0 = V'_{GR}(r) = \frac{d~}{dr}\,V_{GR}(r)= -\frac{2\,h^2}{r^3}
     + \frac{2}{r^2} + \frac{6\,h^2}{r^4}\,,$$
where we used $m_s=1$. Multiplying with $r^4$ gives 
$$ 0 = r^2 - h^2\,r + 3\,h^2 $$
with the solutions
$$r_{\pm}=\frac{h^2}{2}\pm\sqrt{\left(\frac{h^2}{2}\right)^2-3h^2}\,,$$
where $V^{\max}_{GR}=V_{GR}(r_-)$ and $V^{\min}_{GR}=V_{GR}(r_+)$.
Continueing with $h=10^4\,[m_s]$, we find
$$ r_- = 3\,[m_s]~~{\rm  and}~~r_+=10^8\,[m_s]~~~{\rm with}~~~
   V^{\max}_{GR}=3.704\times 10^6~~{\rm and}~~V^{\min}_{GR}=-10^{-8}\,. $$

\item See the first figure (Newton's effective potential $V_N$ is 
included, but was not asked).

\item See the second figure (Newton's effective potential $V_N$ is 
included, but was not asked).

\end{enumerate}


\item{\bf Schwarzschild radius and density (10\%).} 

The Scharzschild radius is 
$$ r_s = 2m = \frac{8\pi}{3}\,r^3\,\rho~\Rightarrow~R_0 =
   \frac{8\pi}{3}\,R_0^3\,\rho~\Rightarrow~R_0^2=\frac{3}{8\pi\,\rho}$$
and $r_S>R$ for $R>R_0$, because $r_s$ increases $\sim R^3$, i.e., faster
than $R$. This rules out a minimum at $r_s=R_0$:
$$ r_s-R = \frac{8\pi}{3}\,\rho\,R^3-R= \frac{8\pi}{3}\,\rho\,R\,
   \left(R^2-\frac{3}{8\pi\,\rho}\right) = \frac{8\pi}{3}\,\rho\,
   R\,\left(R^2-R_0^2\right)\,. $$

\item{\bf Einstein's equations.}

\begin{enumerate}

\item We have
$$ R^{\mu}_{~\mu}-\frac{1}{2}\,g^{\mu}_{~\mu}\,R = 
   -\kappa\,T^{\mu}_{~\mu}~\Rightarrow~R-2\,R = - 
   \kappa\,T~\Rightarrow~R = \kappa\,T$$
and, therefore,
$$ R_{\mu\nu}-\kappa\,\frac{1}{2}\,g_{\mu\nu}\,T = -\kappa\,T_{\mu\nu}
   ~\Rightarrow~ R_{\mu\nu} =-\kappa\,\left(T_{\mu\nu}-
    \frac{1}{2}\,g_{\mu\nu}\,T\right)\,.$$

\item In vacuum we have
$$ R^{\mu}_{~\mu}-\frac{1}{2}\,g^{\mu}_{~\mu}\,R + 
   g^{\mu}_{~\mu}\,\Lambda = 0~\Rightarrow~R-2\,R+4\,\Lambda = 0
   ~\Rightarrow~R=4\,\Lambda\,,$$
i.e., $n=4$.

\end{enumerate} 

\end{enumerate} 

\end{document}
